Question

Two parallel plates are connected to a battery with a potential difference between the plates of 3 Volts. A small charged particle is placed near one of the plates and released. This particle then moves towards the opposite plate, with the electric field doing work on the particle.[Let E_k be kinetic energy, U be potential energy and W the work done].
Which of the following statements is correct? (Give all CORRECT answers: B, AC, BCD ...)

A) The potential energy of the particle increases
B) The kinetic energy of the particle decreases
C) The total energy (kinetic + potential) of the particle increases
D) ?E_k = -W
E) The kinetic energy of the particle increases

Answer 1

option A is wrong because potential energy decreases as charge moves and that decrease in potential energy is converted into kinetic energy. so kinetic energy increases

option B is wrong as explained above

option C is wrong because according to conservation of energy, total energy remain constant.

option D is wrong because delta Ek = W accroding to work energy theorem. so negative sign must not be there.

opton E is correct as explained in option A

so option E is correct